Let X equal the number of typos on a printed page with a mean of 4 typos per page.(a) What is the probability that a randomly selected page has at least one typo on it?(b) What is the probability that a randomly selected page has at most one typo on it?

Answer:a) There is a 98.17% probability that a randomly selected page has at least one typo on it. b) There is a 9.16% probability that a randomly selected page has at most one typo on it.Step-by-step explanation:Since we only have the mean, we can solve this problem by a Poisson distribution.In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]In whichx is the number of sucessese = 2.71828 is the Euler number[tex]\mu[/tex] is the mean in the given time interval.In this problem, we have that [tex]\mu = 4[/tex](a) What is the probability that a randomly selected page has at least one typo on it?Thats is [tex]P(X \geq 1)[/tex]. Either a number is greater or equal than 1, or it is lesser. The sum of the probabilities must be decimal 1. So:[tex]P(X < 1) + P(X \geq 1) = 1[/tex][tex]P(X \geq 1) = 1 - P(X < 1)[/tex]In which[tex]P(X < 1) = P(X = 0)[/tex].So[tex]P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183[/tex][tex]P(X \geq 1) = 1 - P(X < 1) = 1 - 0.0183 = 0.9817[/tex]There is a 98.17% probability that a randomly selected page has at least one typo on it. (b) What is the probability that a randomly selected page has at most one typo on it?This is [tex]P = P(X = 0) + P(X = 1)[/tex]. So:[tex]P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183[/tex][tex]P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733[/tex][tex]P = P(X = 0) + P(X = 1) = 0.0183 + 0.0733 = 0.0916[/tex]There is a 9.16% probability that a randomly selected page has at most one typo on it.